Question

A cone is cut into two parts by a horizontal plane passing through the midpoint of its axis. The ratio of the volume of the upper part to the volume of the lower part is…………………..
$
  (a){\text{ 1:7}} \\
  (b){\text{ 1:8}} \\
  (c){\text{ 7:1}} \\
  (d){\text{ 7:8}} \\
$

Answer 1

Hint: The cone is cut by a horizontal plane into two equal portions thus the height of both the portions must be the same as the cutting point is midpoint (given in the question). Use the direct formula for volume of cone to calculate the volume of top portion, the lower portion formed is a frustum but it’s volume can be calculated by subtracting the overall volume of the main cone with that of top portion of cone which is cut away.

Complete step-by-step answer:

Let us consider the cone as shown in figure the base radius (R) of the cone is R cm (see figure.)
Now the cone cut into two parts through the midpoint of its height (at point B see figure).
Let the height of the cone be h cm.
$ \Rightarrow AB = BC = \dfrac{h}{2}$ cm.
Let the base radius of the smaller cone is r cm.
$ \Rightarrow BE = r$cm (see figure).
Now the triangle ABE and the triangle ACD is concurrent by the property of angle-angle-angle (AAA).
$ \Rightarrow \dfrac{{AB}}{{AC}} = \dfrac{{BE}}{{CD}}$
Now substitute the values in above equation we have,
$ \Rightarrow \dfrac{{\dfrac{h}{2}}}{h} = \dfrac{r}{R}$
On simplifying we get.
$ \Rightarrow r = \dfrac{R}{2}$ cm.
As we know that the volume of the cone is $ = \dfrac{1}{3}\pi {\left( {{\text{radius}}} \right)^2}\left( {{\text{height}}} \right)$
So the volume (${V_1}$) of the upper part of the bigger cone which is also a cone is
$ \Rightarrow {V_1} = \dfrac{1}{3}\pi {\left( {BE} \right)^2}\left( {AB} \right) = \dfrac{1}{3}\pi {\left( r \right)^2}\left( {\dfrac{h}{2}} \right)$…………………. (1)
Let the volume of the lower part is (${V_2}$) so the volume of the lower part is calculated as subtraction of volume of bigger cone and volume of smaller cone.
Volume (V) of bigger cone is $ = \dfrac{1}{3}\pi {\left( {{\text{radius}}} \right)^2}\left( {{\text{height}}} \right)$
$ \Rightarrow V = \dfrac{1}{3}\pi {\left( R \right)^2}h$
So the lower part volume is
$ \Rightarrow {V_2} = V - {V_1}$
$ \Rightarrow {V_2} = \dfrac{1}{3}\pi {\left( R \right)^2}h - \dfrac{1}{3}\pi {\left( r \right)^2}\left( {\dfrac{h}{2}} \right)$
So, the required ratio (RT) of two parts of the cone is
$ \Rightarrow RT = \dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{\dfrac{1}{3}\pi {{\left( r \right)}^2}\left( {\dfrac{h}{2}} \right)}}{{\dfrac{1}{3}\pi {{\left( R \right)}^2}h - \dfrac{1}{3}\pi {{\left( r \right)}^2}\left( {\dfrac{h}{2}} \right)}}$
Now substitute the values in the above equation we have,
$ \Rightarrow RT = \dfrac{{\dfrac{1}{3}\pi {{\left( {\dfrac{R}{2}} \right)}^2}\left( {\dfrac{h}{2}} \right)}}{{\dfrac{1}{3}\pi {{\left( R \right)}^2}h - \dfrac{1}{3}\pi {{\left( {\dfrac{R}{2}} \right)}^2}\left( {\dfrac{h}{2}} \right)}}$
Now cancel out the common term $\dfrac{1}{3}\pi \left( {\dfrac{h}{2}} \right)$ from, numerator and denominator and simplify we have,
$ \Rightarrow RT = \dfrac{{{{\left( {\dfrac{R}{2}} \right)}^2}}}{{\left[ {2{{\left( R \right)}^2} - {{\left( {\dfrac{R}{2}} \right)}^2}} \right]}} = \dfrac{{\dfrac{1}{4}}}{{\left( {2 - \dfrac{1}{4}} \right)}} = \dfrac{{\dfrac{1}{4}}}{{\dfrac{7}{4}}} = \dfrac{1}{7}$
So, this is the required ratio of the two parts of the cone.
Hence option (A) is correct.

Note: Whenever we face such types of problems the key concept is to have a good understanding of the direct formula for volume of some basic conic sections like cone, frustum, sphere, cylinder etc. The volume of the lower portion could be calculated by an alternate method which is to consider the lower portion as a frustum. The formula for volume of frustum is $\dfrac{1}{3}\pi h\left( {{R_1}^2 + {R_2}^2 + {R_1}{R_2}} \right)$.